### Sequence and Series : Notes and Study Materials -pdf

**Sequence**

It means an arrangement of number in definite order according to some rule

$2,4,6,8,10, 12,14………$

$1, 3, 5,9, 11, 13……..$

$1^2 +2^2 + 3^3 +……….$

Important point

- The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
- Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
- The nth term is the number at the nth position of the sequence and is denoted by a
_{n} - The nth term is also called the general term of the sequence
- A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3…k}.
- Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme

1. 0,2, 4,6,8…….

General Term a

_{n}= 2n

2. 1, 1, 2, 3, 5, 8,..

This sequence does not have visible algebraic formula, but the sequence is generated by the recurrence relation given by

a

_{1}=a

_{2}=1

an =a

_{n-2}+a

_{n-1}, n > 2

This sequence is called Fibonacci sequence

**Finite sequence**

A sequence containing a finite number of terms is called Finite sequence**infinite sequence**

A sequence is called infinite if it is not a finite sequence.

**Series**

Let a_{1} , a_{2} , a_{3} ..be the sequence, then the sum expressed a_{1} + a_{2} +a_{3} + ……. is called series.

- A series is called finite series if it has got finite number of terms
- A series is called infinite series if it has got infinite terms
- Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)
- so , a
_{1}+ a_{2}+a_{3}+ …….a_{n}can be expressed as $\sum_{k=1}^{n}a_k$

Example

$1+2+3 + 4+ 5+….+100$

$= \sum_{k=1}^{n}k$

### Types of Sequences

### Arithmetic Progression

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant

Examples

1. $1,5,9,13,17….$

2. $1,2,3,4,5,…..$__Important Notes about Arithmetic Progression__

1.The difference between any successive members is a constant and it is called the common difference of AP

2. If a_{1}, a_{2},a_{3},a_{4},a_{5} are the terms in AP then

D=a_{2 }-a_{1} =a_{3} – a_{2} =a_{4} – a_{3}=a_{5} –a_{4}

3. We can represent the general form of AP in the form

a,a+d,a+2d,a+3d,a+4d..

Where a is first term and d is the common difference

4. Nth term of Arithmetic Progression is given by n^{th}term = a + (n – 1)d

5. Sum of nth item in Arithmetic Progression is given by

$S_n =\frac {n}{2}[a + (n-1)d]$

Or

$S_n =\frac {n}{2}[t_1+ t_n]$__Arithmetic Mean__

The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP

A -a = b- A or $A = \frac {(a+b)}{2}$

If we want to add n terms between A and B so that result sequence is in AP

Let A_{1}, A_{2} , A_{3} , A_{4} , A_{5} …. A_{n} be the terms added between a and b

Then

b= a+ [(n+2) -1]d

or $d = \frac {(b-a)}{(n+1)}$

So terms will be $a+ \frac {(b-a)}{(n+1)}, a + 2 \frac {(b-a)}{(n+1)} ,……. a+ n \frac {(b-a)}{(n+1)}$

### Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout.

Examples

1. $2,4,8,16,32….$

2. $3,6,12,24,48,…..$__Important Notes about Geometric Progressions__

- The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.
- If a
_{1}, a_{2},a_{3},a_{4},a_{5}are the terms in AP then

$r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$ - We can represent the general form of G.P in the form

a,ar,ar^{2},ar^{3}……

Where a is first term and r is the common ratio - Nth term of Geometric Progression is given by n
^{th}term = ar^{n-1} - Sum of nth item inGeometric Progression is given by $S_n =a \frac {r^n -1}{r-1}$

Or

$S_n =a \frac {1-r^n }{ 1-r}$

__Geometric Means__

The Geometric mean G of any two numbers a and b is given by (ab)^{1/2} i.e. a, G, b are in G.P

$\frac {G}{a} = \frac {b}{G}$

or $G=\sqrt {ab}$

If we want to add n terms between A and B so that result sequence is in GP

Let G_{1}, G_{2} , G_{3} , G_{4} , G_{5} …. G_{n} be the terms added between a and b

Then

$b= ar^{n+1}$

or $r= (\frac {b}{a})^{\frac {1}{n+1}}$

So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,……. a(\frac {b}{a})^{\frac {n}{n+1}}$

Relationship Between A.P and G.P

Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.

Then

A ≥ G

And

$A – G = \frac {{\sqrt {a} – \sqrt {b}}^2}{2}$

### Sum of Special Series

__Sum of first n natural numbers__

$1 + 2 + 3 +….. + n = \frac {n(n+1}{2}$

$ \sum_{k=1}^{n} n =\frac {n(n+1}{2}$__sum of squares of the first n natural numbers__

$1^2+ 2^2+3^2 +…. + n^2 = \frac {n(n+1)(2n+1)}{6}$

$ \sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$__Sum of cubes of the first n natural numbers__

$1^3+ 2^3+3^3 +….. + n^3 = \frac {(n(n+1))^2}{4}$

$ \sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$

### Rules for finding the sum of Series

a. Write the nth term T_{n}of the series

b. Write the T_{n} in the polynomial form of n

$T_n= a n^3 + bn^2 + cn +d$

c. The sum of series can be written as

$ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd

We already know the values if these standard from the formula given above and we can easily find the sum of the series

Example

Find the sum of the series

$2^2 + 4^2 + 6^2 + …..(2n)^2}

Solution

Let nth term T_{n}of the series

Then

$T_n = (2n)^2 = 4n^2$

Now

$2^2 + 4^2 + 6^2 + …..(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$

$=\frac {2n(n+1)(2n+1)}{3}$

### Method of Difference

Many times , nth term of the series can be determined. For example

5 + 11 + 19 + 29 + 41……

If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference

$S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$

or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$

On subtraction, we get

$0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$

Here 6,8,10 is in A.P,So

$a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$

or $ a_n= n^2 + 3n + 1$

Now it is easy to find the Sum of the series

$S_n = \sum_{k=1}^{n} {k^2 +3k +1}$

$=\frac {n(n+2)(n+4)}{3}$

### Sequences and Series Class 11 MCQs Questions with Answers

Question 1.

If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in

(a) AP

(b) GP

(c) HP

(d) none of these

## Answer

Answer: (a) AP

Hint:

Given a, b, c are in GP

⇒ b² = ac

⇒ b² – ac = 0

So, ax² + 2bx + c = 0 have equal roots.

Now D = 4b² – 4ac

and the root is -2b/2a = -b/a

So -b/a is the common root.

Now,

dx² + 2ex + f = 0

⇒ d(-b/a)² + 2e×(-b/a) + f = 0

⇒ db2 /a² – 2be/a + f = 0

⇒ d×ac /a² – 2be/a + f = 0

⇒ dc/a – 2be/a + f = 0

⇒ d/a – 2be/ac + f/c = 0

⇒ d/a + f/c = 2be/ac

⇒ d/a + f/c = 2be/b²

⇒ d/a + f/c = 2e/b

⇒ d/a, e/b, f/c are in AP

Question 2.

If a, b, c are in AP then

(a) b = a + c

(b) 2b = a + c

(c) b² = a + c

(d) 2b² = a + c

## Answer

Answer: (b) 2b = a + c

Hint:

Given, a, b, c are in AP

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

Question 3:

Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is

(a) 2 + √3

(b) 2 – √3

(c) 2 ± √3

(d) None of these

## Answer

Answer: (a) 2 + √3

Hint:

Let the three numbers be a/r, a, ar

Since the numbers form an increasing GP, So r > 1

Now, it is given that a/r, 2a, ar are in AP

⇒ 4a = a/r + ar

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± √3

⇒ r = 2 + √3 {Since r > 1}

Question 4:

The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

(a) n/(n+1)

(b) 1/(n+1)

(c) 1/n

(d) None of these

## Answer

Answer: (a) n/(n+1)

Hint:

Given series is:

S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

Question 5:

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP

(b) a², b², c² are in AP

(c) 1/1, 1/b, 1/c are in AP

(d) None of these

## Answer

Answer: (b) a², b², c² are in AP

Hint:

Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

Question 6:

The sum of series 1/2! + 1/4! + 1/6! + ….. is

(a) e² – 1 / 2

(b) (e – 1)² /2 e

(c) e² – 1 / 2 e

(d) e² – 2 / e

## Answer

Answer: (b) (e – 1)² /2 e

Hint:

We know that,

e^{x} = 1 + x/1! + x² /2! + x³ /3! + x^{4} /4! + ………..

Now,

e^{1} = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..

e^{-1} = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..

e^{1} + e^{-1} = 2(1 + 1/2! + 1/4! + ………..)

⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..

⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..

⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..

⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7:

The third term of a geometric progression is 4. The product of the first five terms is

(a) 4^{3}

(b) 4^{5}

(c) 4^{4}

(d) none of these

## Answer

Answer: (b) 4^{5}

Hint:

here it is given that T_{3} = 4.

⇒ ar² = 4

Now product of first five terms = a.ar.ar².ar³.ar^{4}

= a^{5}r^{10}

= (ar^{2})^{5}

= 4^{5}

Question 8:

Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals

(a) 1/m n

(b) 1/m + 1/n

(c) 1

(d) 0

## Answer

Answer: (c) 1

Hint:

Let first term is a and the common difference is d of the AP

Now, T_{m} = 1/n

⇒ a + (m-1)d = 1/n ………… 1

and T_{n} = 1/m

⇒ a + (n-1)d = 1/m ………. 2

From equation 2 – 1, we get

(m-1)d – (n-1)d = 1/n – 1/m

⇒ (m-n)d = (m-n)/mn

⇒ d = 1/mn

From equation 1, we get

a + (m-1)/mn = 1/n

⇒ a = 1/n – (m-1)/mn

⇒ a = {m – (m-1)}/mn

⇒ a = {m – m + 1)}/mn

⇒ a = 1/mn

Now, T_{mn} = 1/mn + (mn-1)/mn

⇒ T_{mn} = 1/mn + 1 – 1/mn

⇒ T_{mn} = 1

Question 9.

The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Answer: (c) 6

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 10.

If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in

(a) A.P.

(b) G.P.

(c) H.P.

(d) A.G.P.

## Answer

Answer: (c) H.P.

Hint:

Given, equation is

ax² + bx + c = 0

Let p and q are the roots of this equation.

Now p+q = -b/a

and pq = c/a

Given that

p + q = 1/p² + 1/q²

⇒ p + q = (p² + q²)/(p² ×q²)

⇒ p + q = {(p + q)² – 2pq}/(pq)²

⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²

⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}

⇒ -bc²/a³ = {b² – 2ca}/a²

⇒ -bc²/a = b² – 2ca

Divide by bc on both side, we get

⇒ -c /a = b/c – 2a/b

⇒ 2a/b = b/c + c/a

⇒ b/c, a/b, c/a are in AP

⇒ c/a, a/b, b/c are in AP

⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP

⇒ a/c, b/a, c/b are in HP

Question 11.

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP

(b) a², b², c² are in AP

(c) 1/1, 1/b, 1/c are in AP

(d) None of these

## Answer

Answer: (b) a², b², c² are in AP

Hint:

Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

Question 12.

The 35th partial sum of the arithmetic sequence with terms a_{n} = n/2 + 1

(a) 240

(b) 280

(c) 330

(d) 350

## Answer

Answer: (d) 350

Hint:

The 35th partial sum of this sequence is the sum of the first thirty-five terms.

The first few terms of the sequence are:

a_{1} = 1/2 + 1 = 3/2

a_{2} = 2/2 + 1 = 2

a_{3} = 3/2 + 1 = 5/2

Here common difference d = 2 – 3/2 = 1/2

Now, a_{35} = a_{1} + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2

Now, the sum = (35/2) × (3/2 + 37/2)

= (35/2) × (40/2)

= (35/2) × 20

= 35 × 10

= 350

Question 13.

The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Answer: (c) 6

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 14.

The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

Answer: (c) 3

Hint:

Let first term of the GP is a and common ratio is r.

3rd term = ar²

5th term = ar^{4}

Now

⇒ ar² + ar^{4} = 90

⇒ a(r² + r^{4}) = 90

⇒ r² + r^{4} = 90

⇒ r² ×(r² + 1) = 90

⇒ r²(r² + 1) = 3² ×(3² + 1)

⇒ r = 3

So the common ratio is 3

Question 15.

The sum of AP 2, 5, 8, …..up to 50 terms is

(a) 3557

(b) 3775

(c) 3757

(d) 3575

## Answer

Answer: (b) 3775

Hint:

Given, AP is 2, 5, 8, …..up to 50

Now, first term a = 2

common difference d = 5 – 2 = 3

Number of terms = 50

Now, Sum = (n/2)×{2a + (n – 1)d}

= (50/2)×{2×2 + (50 – 1)3}

= 25×{4 + 49×3}

= 25×(4 + 147)

= 25 × 151

= 3775

Question 16.

If 2/3, k, 5/8 are in AP then the value of k is

(a) 31/24

(b) 31/48

(c) 24/31

(d) 48/31

## Answer

Answer: (b) 31/48

Hint:

Given, 2/3, k, 5/8 are in AP

⇒ 2k = 2/3 + 5/8

⇒ 2k = 31/24

⇒ k = 31/48

So, the value of k is 31/48

Question 17.

The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

(a) n/(n+1)

(b) 1/(n+1)

(c) 1/n

(d) None of these

## Answer

Answer: (a) n/(n+1)

Hint:

Given series is:

S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

Question 18.

If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is

(a) 228

(b) 74

(c) 740

(d) 1090

## Answer

Answer: (c) 740

Hint:

Let a is the first term and d is the common difference of AP

Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term

⇒ a + 2d = 7 ………….. 1

and

3(a + 2d) + 2 = a + 6d

⇒ 3×7 + 2 = a + 6d

⇒ 21 + 2 = a + 6d

⇒ a + 6d = 23 ………….. 2

From equation 1 – 2, we get

4d = 16

⇒ d = 16/4

⇒ d = 4

From equation 1, we get

a + 2×4 = 7

⇒ a + 8 = 7

⇒ a = -1

Now, the sum of its first 20 terms

= (20/2)×{2×(-1) + (20-1)×4}

= 10×{-2 + 19×4)}

= 10×{-2 + 76)}

= 10 × 74

= 740

Question 19.

If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals

(a) 10

(b) 12

(c) 11

(d) 13

## Answer

Answer: (c) 11

Hint:

Given,

the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….

⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}

⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}

⇒ 6n + 1 = {2n + 112}/2

⇒ 6n + 1 = n + 56

⇒ 6n – n = 56 – 1

⇒ 5n = 55

⇒ n = 55/5

⇒ n = 11

Question 20.

If a is the A.M. of b and c and G_{1} and G_{2} are two GM between them then the sum of their cubes is

(a) abc

(b) 2abc

(c) 3abc

(d) 4abc

## Answer

Answer: (b) 2abc

Hint:

Given, a is the A.M. of b and c

⇒ a = (b + c)

⇒ 2a = b + c ………… 1

Again, given G_{1} and G_{1} are two GM between b and c,

⇒ b, G_{1}, G_{2}, c are in the GP having common ration r, then

⇒ r = (c/b)^{1/(2+1)} = (c/b)^{1/3}

Now,

G_{1} = br = b×(c/b)^{1/3}

and G_{1} = br = b×(c/b)^{2/3}

Now,

(G_{1})³ + (G_{2})3 = b³ ×(c/b) + b³ ×(c/b)²

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( 1 + c/b)

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( b + c)/b

⇒ (G_{1})³ + (G_{2})³ = b² ×c×( b + c)/b

⇒ (G_{1})³ + (G_{2})³ = b² ×c×( b + c)/b ………….. 2

From equation 1

2a = b + c

⇒ 2a/b = (b + c)/b

Put value of(b + c)/b in eqaution 2, we get

(G_{1})³ + (G_{2})³ = b² × c × (2a/b)

⇒ (G_{1})³ + (G_{2})³ = b × c × 2a

⇒ (G_{1})³ + (G_{2})³ = 2abc

### Sequences and Series Class 11 MCQs Questions with Answers

Question 1.

If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in

(a) AP

(b) GP

(c) HP

(d) none of these

## Answer

Answer: (a) AP

Hint:

Given a, b, c are in GP

⇒ b² = ac

⇒ b² – ac = 0

So, ax² + 2bx + c = 0 have equal roots.

Now D = 4b² – 4ac

and the root is -2b/2a = -b/a

So -b/a is the common root.

Now,

dx² + 2ex + f = 0

⇒ d(-b/a)² + 2e×(-b/a) + f = 0

⇒ db2 /a² – 2be/a + f = 0

⇒ d×ac /a² – 2be/a + f = 0

⇒ dc/a – 2be/a + f = 0

⇒ d/a – 2be/ac + f/c = 0

⇒ d/a + f/c = 2be/ac

⇒ d/a + f/c = 2be/b²

⇒ d/a + f/c = 2e/b

⇒ d/a, e/b, f/c are in AP

Question 2.

If a, b, c are in AP then

(a) b = a + c

(b) 2b = a + c

(c) b² = a + c

(d) 2b² = a + c

## Answer

Answer: (b) 2b = a + c

Hint:

Given, a, b, c are in AP

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

Question 3:

Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is

(a) 2 + √3

(b) 2 – √3

(c) 2 ± √3

(d) None of these

## Answer

Answer: (a) 2 + √3

Hint:

Let the three numbers be a/r, a, ar

Since the numbers form an increasing GP, So r > 1

Now, it is given that a/r, 2a, ar are in AP

⇒ 4a = a/r + ar

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± √3

⇒ r = 2 + √3 {Since r > 1}

Question 4:

The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

(a) n/(n+1)

(b) 1/(n+1)

(c) 1/n

(d) None of these

## Answer

Answer: (a) n/(n+1)

Hint:

Given series is:

S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

Question 5:

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP

(b) a², b², c² are in AP

(c) 1/1, 1/b, 1/c are in AP

(d) None of these

## Answer

Answer: (b) a², b², c² are in AP

Hint:

Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

Question 6:

The sum of series 1/2! + 1/4! + 1/6! + ….. is

(a) e² – 1 / 2

(b) (e – 1)² /2 e

(c) e² – 1 / 2 e

(d) e² – 2 / e

## Answer

Answer: (b) (e – 1)² /2 e

Hint:

We know that,

e^{x} = 1 + x/1! + x² /2! + x³ /3! + x^{4} /4! + ………..

Now,

e^{1} = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..

e^{-1} = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..

e^{1} + e^{-1} = 2(1 + 1/2! + 1/4! + ………..)

⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..

⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..

⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..

⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7:

The third term of a geometric progression is 4. The product of the first five terms is

(a) 4^{3}

(b) 4^{5}

(c) 4^{4}

(d) none of these

## Answer

Answer: (b) 4^{5}

Hint:

here it is given that T_{3} = 4.

⇒ ar² = 4

Now product of first five terms = a.ar.ar².ar³.ar^{4}

= a^{5}r^{10}

= (ar^{2})^{5}

= 4^{5}

Question 8:

Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals

(a) 1/m n

(b) 1/m + 1/n

(c) 1

(d) 0

## Answer

Answer: (c) 1

Hint:

Let first term is a and the common difference is d of the AP

Now, T_{m} = 1/n

⇒ a + (m-1)d = 1/n ………… 1

and T_{n} = 1/m

⇒ a + (n-1)d = 1/m ………. 2

From equation 2 – 1, we get

(m-1)d – (n-1)d = 1/n – 1/m

⇒ (m-n)d = (m-n)/mn

⇒ d = 1/mn

From equation 1, we get

a + (m-1)/mn = 1/n

⇒ a = 1/n – (m-1)/mn

⇒ a = {m – (m-1)}/mn

⇒ a = {m – m + 1)}/mn

⇒ a = 1/mn

Now, T_{mn} = 1/mn + (mn-1)/mn

⇒ T_{mn} = 1/mn + 1 – 1/mn

⇒ T_{mn} = 1

Question 9.

The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Answer: (c) 6

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 10.

If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in

(a) A.P.

(b) G.P.

(c) H.P.

(d) A.G.P.

## Answer

Answer: (c) H.P.

Hint:

Given, equation is

ax² + bx + c = 0

Let p and q are the roots of this equation.

Now p+q = -b/a

and pq = c/a

Given that

p + q = 1/p² + 1/q²

⇒ p + q = (p² + q²)/(p² ×q²)

⇒ p + q = {(p + q)² – 2pq}/(pq)²

⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²

⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}

⇒ -bc²/a³ = {b² – 2ca}/a²

⇒ -bc²/a = b² – 2ca

Divide by bc on both side, we get

⇒ -c /a = b/c – 2a/b

⇒ 2a/b = b/c + c/a

⇒ b/c, a/b, c/a are in AP

⇒ c/a, a/b, b/c are in AP

⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP

⇒ a/c, b/a, c/b are in HP

Question 11.

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP

(b) a², b², c² are in AP

(c) 1/1, 1/b, 1/c are in AP

(d) None of these

## Answer

Hint:

Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

Question 12.

The 35th partial sum of the arithmetic sequence with terms a_{n} = n/2 + 1

(a) 240

(b) 280

(c) 330

(d) 350

## Answer

Answer: (d) 350

Hint:

The 35th partial sum of this sequence is the sum of the first thirty-five terms.

The first few terms of the sequence are:

a_{1} = 1/2 + 1 = 3/2

a_{2} = 2/2 + 1 = 2

a_{3} = 3/2 + 1 = 5/2

Here common difference d = 2 – 3/2 = 1/2

Now, a_{35} = a_{1} + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2

Now, the sum = (35/2) × (3/2 + 37/2)

= (35/2) × (40/2)

= (35/2) × 20

= 35 × 10

= 350

Question 13.

The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is

(a) 2

(b) 4

(c) 6

(d) 8

## Answer

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 14.

The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

Answer: (c) 3

Hint:

Let first term of the GP is a and common ratio is r.

3rd term = ar²

5th term = ar^{4}

Now

⇒ ar² + ar^{4} = 90

⇒ a(r² + r^{4}) = 90

⇒ r² + r^{4} = 90

⇒ r² ×(r² + 1) = 90

⇒ r²(r² + 1) = 3² ×(3² + 1)

⇒ r = 3

So the common ratio is 3

Question 15.

The sum of AP 2, 5, 8, …..up to 50 terms is

(a) 3557

(b) 3775

(c) 3757

(d) 3575

## Answer

Answer: (b) 3775

Hint:

Given, AP is 2, 5, 8, …..up to 50

Now, first term a = 2

common difference d = 5 – 2 = 3

Number of terms = 50

Now, Sum = (n/2)×{2a + (n – 1)d}

= (50/2)×{2×2 + (50 – 1)3}

= 25×{4 + 49×3}

= 25×(4 + 147)

= 25 × 151

= 3775

Question 16.

If 2/3, k, 5/8 are in AP then the value of k is

(a) 31/24

(b) 31/48

(c) 24/31

(d) 48/31

## Answer

Answer: (b) 31/48

Hint:

Given, 2/3, k, 5/8 are in AP

⇒ 2k = 2/3 + 5/8

⇒ 2k = 31/24

⇒ k = 31/48

So, the value of k is 31/48

Question 17.

The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

(a) n/(n+1)

(b) 1/(n+1)

(c) 1/n

(d) None of these

## Answer

Answer: (a) n/(n+1)

Hint:

Given series is:

S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

Question 18.

If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is

(a) 228

(b) 74

(c) 740

(d) 1090

## Answer

Answer: (c) 740

Hint:

Let a is the first term and d is the common difference of AP

Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term

⇒ a + 2d = 7 ………….. 1

and

3(a + 2d) + 2 = a + 6d

⇒ 3×7 + 2 = a + 6d

⇒ 21 + 2 = a + 6d

⇒ a + 6d = 23 ………….. 2

From equation 1 – 2, we get

4d = 16

⇒ d = 16/4

⇒ d = 4

From equation 1, we get

a + 2×4 = 7

⇒ a + 8 = 7

⇒ a = -1

Now, the sum of its first 20 terms

= (20/2)×{2×(-1) + (20-1)×4}

= 10×{-2 + 19×4)}

= 10×{-2 + 76)}

= 10 × 74

= 740

Question 19.

If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals

(a) 10

(b) 12

(c) 11

(d) 13

## Answer

Answer: (c) 11

Hint:

Given,

the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….

⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}

⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}

⇒ 6n + 1 = {2n + 112}/2

⇒ 6n + 1 = n + 56

⇒ 6n – n = 56 – 1

⇒ 5n = 55

⇒ n = 55/5

⇒ n = 11

Question 20.

If a is the A.M. of b and c and G_{1} and G_{2} are two GM between them then the sum of their cubes is

(a) abc

(b) 2abc

(c) 3abc

(d) 4abc

## Answer

Answer: (b) 2abc

Hint:

Given, a is the A.M. of b and c

⇒ a = (b + c)

⇒ 2a = b + c ………… 1

Again, given G_{1} and G_{1} are two GM between b and c,

⇒ b, G_{1}, G_{2}, c are in the GP having common ration r, then

⇒ r = (c/b)^{1/(2+1)} = (c/b)^{1/3}

Now,

G_{1} = br = b×(c/b)^{1/3}

and G_{1} = br = b×(c/b)^{2/3}

Now,

(G_{1})³ + (G_{2})3 = b³ ×(c/b) + b³ ×(c/b)²

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( 1 + c/b)

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( b + c)/b

⇒ (G_{1})³ + (G_{2})³ = b² ×c×( b + c)/b

⇒ (G_{1})³ + (G_{2})³ = b² ×c×( b + c)/b ………….. 2

From equation 1

2a = b + c

⇒ 2a/b = (b + c)/b

Put value of(b + c)/b in eqaution 2, we get

(G_{1})³ + (G_{2})³ = b² × c × (2a/b)

⇒ (G_{1})³ + (G_{2})³ = b × c × 2a

⇒ (G_{1})³ + (G_{2})³ = 2abc